advanced32 min studyLesson 515 in Mathematics

Regular representations and group algebras

Use the regular character to count every irreducible and identify the matrix blocks inside the complex group algebra.

On this page

What you’ll understand

  • Calculate the left regular character and its multiplicities.
  • Interpret the sum-of-squares dimension identity.
  • Construct central idempotents and distinguish irreducible from isotypic summands.

One representation contains every complex irreducible of a finite group: let the group translate its own elements and extend that permutation action linearly. This regular representation is large but exceptionally easy to analyze because nonidentity translations fix no basis label. Its character yields a dimension identity that checks the completeness of any proposed irreducible list. Extending multiplication linearly then turns the same vector space into the group algebra, whose decomposition into matrix blocks explains why Fourier transforms will eventually be matrix valued.

Translation on the group itself

The complex group algebra has one basis symbol for each group element. Multiply basis symbols using the group law, then distribute over sums and scalar coefficients. The identity basis symbol is the multiplicative identity. Left multiplication by a group element permutes this basis and defines the left regular representation. A nonidentity element cannot fix any basis label, since cancelling that label would force the translating element to be the identity. The identity operator fixes the entire basis.

Consequently the regular character is the group order at the identity and zero elsewhere. Its inner product with an irreducible character is the dimension of that irreducible. This proves every irreducible occurs, rather than assuming that a previously found list is complete. Comparing dimensions gives the sum of squared irreducible dimensions equal to the group order. Counting dimensions only once would miss the multiplicity built into the regular representation; a two-dimensional irreducible contributes four dimensions here.

χreg(g)={∣G∣,g=e,0,g≠e,C[G]≅⨁ρ∈G^Vρ⊕dρ,∣G∣=∑ρ∈G^dρ2\chi_{\mathrm{reg}}(g)=\begin{cases}|G|,&g=e,\\0,&g\ne e,\end{cases}\qquad \mathbb C[G]\cong\bigoplus_{\rho\in\widehat G}V_\rho^{\oplus d_\rho},\qquad |G|=\sum_{\rho\in\widehat G}d_\rho^2
The set of inequivalent complex irreducibles is denoted by the hat on G, and dρ is the dimension of the corresponding irreducible space.↔ Scroll wide equations horizontally.

From a module decomposition to algebra blocks

Every representation extends linearly to an action of the group algebra: a linear combination of group elements acts by the same linear combination of their matrices. Applying all inequivalent irreducible representations at once gives an algebra homomorphism into a direct sum of full endomorphism algebras. Over the complex numbers it is an isomorphism. One route uses semisimplicity and the structure theorem for finite-dimensional semisimple complex algebras; the regular multiplicities then identify the sizes of its matrix blocks.

The field restriction is significant. Maschke's complete reducibility works over any field with invertible group order, but the simple algebra blocks over a general field can involve nontrivial division algebras rather than just matrices over that field. In characteristic dividing the order, even semisimplicity can fail. We therefore state the full matrix-block decomposition over the complex numbers here. It says the group algebra has exactly enough independent matrix entries to encode an arbitrary function on the finite group, anticipating Fourier inversion.

C[G]⟶⨁ρ∈G^End⁡C(Vρ),∑gagg⟼(∑gagρ(g))ρ\mathbb C[G]\longrightarrow\bigoplus_{\rho\in\widehat G}\operatorname{End}_{\mathbb C}(V_\rho),\qquad \sum_g a_g g\longmapsto\left(\sum_g a_g\rho(g)\right)_\rho
This is an isomorphism of complex algebras. Multiplication on the target is matrix multiplication separately in each irreducible block.↔ Scroll wide equations horizontally.

Central projectors isolate types

A central idempotent is an algebra element whose square equals itself and which commutes with every algebra element. A primitive central idempotent associated with one irreducible type projects onto its invariant isotypic component: the sum of all copies of that type. An arbitrary central idempotent can select several whole isotypic components at once. Even a primitive one does not isolate a single copy when that type has multiplicity greater than one. For the trivial type, averaging all group elements gives the projection onto fixed vectors. Multiplying by another average leaves it unchanged because every group element appears equally often.

For a general irreducible type, the coefficient of each group element in its central idempotent is the dimension times the character of the inverse element divided by the group order. Centrality follows from the character being constant on conjugacy classes. Schur's lemma makes its action scalar on each irreducible; taking traces and applying character orthogonality makes that scalar one for the selected type and zero for all others. Distinct such idempotents multiply to zero, and their sum is the identity.

Find the three central blocks of the S3 regular action

Use irreducible dimensions 1,1,21,1,2 and standard character (2,0,−1)(2,0,-1) to describe the regular representation and the central projector onto its standard isotypic part.

  1. The regular multiplicities equal dimensions, so C[S3]≅1⊕ε⊕U⊕U\mathbb C[S_3]\cong\mathbf1\oplus\varepsilon\oplus U\oplus U. The dimension check is 1+1+2+2=61+1+2+2=6.

  2. The standard central idempotent is eU=26∑gχU(g−1)g=23e−13[(123)+(132)]e_U=\tfrac{2}{6}\sum_g\chi_U(g^{-1})g=\tfrac23 e-\tfrac13[(123)+(132)]. Transpositions have coefficient zero.

  3. Let c=(123)+(132)c=(123)+(132). Then c2=c+2ec^2=c+2e, so squaring (2e−c)/3(2e-c)/3 gives (4e−4c+c2)/9=(2e−c)/3(4e-4c+c^2)/9=(2e-c)/3. Its trace in the regular action is six times its identity coefficient, namely four.

The result

The central projector has a four-dimensional image consisting of two standard copies. The algebra decomposition is two scalar blocks and one two-by-two matrix block.

Explore the model

Interactive lab

Interactive lab · modular

A regular cyclic translation has a full orbit

Pause and Reset and set both modulus m=6m=6 and step a=1a=1. Predict six distinct residues before the return to zero; Step advances one addition each half-second. Reset and set both m=6,a=2m=6,a=2, predicting a three-residue orbit. Both nonidentity translations fix no residue, despite different orbit lengths. The diagram follows only the orbit from zero and does not display group-algebra coefficients or central projectors. Changing controls preserves time; use Reset for each comparison.

A regular cyclic translation has a full orbitAddition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Current readings: Residue r0: 0; Orbit period · m/gcd(m,a): 12; Distinct residues visited: 1 of 12. All parameters are adjustable with the labeled controls below.Repeated addition modulo 12: the current residue is 0, and the orbit period is 12.01234567891011+ 5mod 12An additive orbitn = 0rₙ = 0Period = 12Cycle starts at 0:0 → 5 → 10 → 3 →8 → 1 → 6 → 11 →4 → 9 → 2 → 7 → 0Exact residues · two additions per second
Additive orbitCurrent residue / next edge
Residue r0
0
Orbit period · m/gcd(m,a)
12
Distinct residues visited
1 of 12
t = 0 s

Addition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forRegular-character values count fixed basis labels, which is different from the length of one orbit.

Check your understanding

A complex irreducible representation has dimension three. How many dimensions does its isotypic component occupy in the regular representation?

Try it yourself

For C4=⟨r⟩C_4=\langle r\rangle, let χk(rj)=ikj\chi_k(r^j)=i^{kj} for k=0,1,2,3k=0,1,2,3. Decompose its regular representation and write the central idempotent selecting the character with k=1k=1. Check how r acts on that idempotent.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

All irreducibles are one dimensional. Use inverse-element character coefficients, then shift the index after left multiplication by r.

Compare with the solution

The regular representation is the sum of the four distinct characters once each. The desired idempotent is e1=(e−ir−r2+ir3)/4e_1=(e-ir-r^2+ir^3)/4. Multiplication gives re1=(r−ir2−r3+ie)/4=ie1re_1=(r-ir^2-r^3+ie)/4=ie_1, so its image is the line on which r acts by i. The coefficient signs use the inverse character, not the unmodified row; reversing that choice selects the conjugate character instead.

Take it with you

  • Regular multiplicities equal irreducible dimensions over the complex numbers.
  • The sum of squared dimensions checks a complete irreducible list.
  • Central idempotents select whole isotypic components, including every repeated copy.

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